2017-04-27 34 views
0

我有以下代碼中我想獲得騎手的id,但我不知道如何做到這一點,其實騎手是鏈接的用戶類型到RiderLocation表,當我改變騎手用戶的類型,然後我不能發送的ID在URL的參數這樣如何獲得JPA存儲庫中的鏈接表的ID

http://localhost:3000/api/riderLocations/search/findByRider?rider_id=3

代碼如下

RiderLocation.java

@Entity 
public class RiderLocation implements Serializable { 

    private static final long serialVersionUID = 1L; 
     @Id 
     @GeneratedValue(strategy = GenerationType.IDENTITY) 
     private Long id; 

     @ManyToOne 
     private User rider; 

     private Double latitude; 
     private Double longitude; 

     public Long getId() { 
      return id; 
     } 

     public void setId(Long id) { 
      this.id = id; 
     } 
     public User getRider() { 
      return rider; 
     } 
     public void setRider(User rider) { 
      this.rider = rider; 
     } 
     public Double getLatitude() { 
      return latitude; 
     } 

     public void setLatitude(Double latitude) { 
      this.latitude = latitude; 
     } 

     public Double getLongitude() { 
      return longitude; 
     } 
     public void setLongitude(Double longitude) { 
      this.longitude = longitude; 
     } 
    } 

RiderLocationRepository

public interface RiderLocationRepository extends JpaRepository<RiderLocation, Long>{ 

    @Query("select r.latitude,r.longitude from RiderLocation r ") 
    ArrayList<Object[]> findAllRidersLocation(); 
    RiderLocation findByRider(@Param("rider_id") Long rider); 

} 
+0

也是騎士對象的一部分? – sAm

+0

'id'是RiderLOcation id @sAm的一部分 –

回答

0

的問題現在已經解決了 唯一的錯誤是我寫的

findByRider(@Param("rider_id") Long rider);

代替:

findByRiderId(@Param("rider_id") Long rider);

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